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JEE Advanced 2007 · Vector Algebra

Let a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c} be unit vectors such that a⃗+b⃗+c⃗=0⃗\vec{a}+\vec{b}+\vec{c}=\vec{0}. Which one of the following is correct ?

Answer: B. a⃗×b⃗=b⃗×c⃗=c⃗×a⃗≠0⃗\vec{a}\times\vec{b} = \vec{b}\times\vec{c} = \vec{c}\times\vec{a} \neq \vec{0}

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Written solution

Given: ∣a⃗∣=∣b⃗∣=∣c⃗∣=1\text{Given: } |\vec{a}| = |\vec{b}| = |\vec{c}| = 1

a⃗+b⃗+c⃗=0⃗\vec{a}+\vec{b}+\vec{c}=\vec{0}

To find: relation between a⃗×b⃗, b⃗×c⃗, c⃗×a⃗\text{To find: relation between } \vec{a}\times\vec{b},\ \vec{b}\times\vec{c},\ \vec{c}\times\vec{a}

∣a⃗∣=∣b⃗∣=∣c⃗∣=1|\vec{a}|=|\vec{b}|=|\vec{c}|=1

∠(a⃗,b⃗)=∠(b⃗,c⃗)=∠(c⃗,a⃗)=120∘\angle(\vec{a},\vec{b}) = \angle(\vec{b},\vec{c}) = \angle(\vec{c},\vec{a}) = 120^\circ

Plan: cross a⃗+b⃗+c⃗=0⃗ with each of a⃗, b⃗, c⃗\text{Plan: cross } \vec{a}+\vec{b}+\vec{c}=\vec{0} \text{ with each of } \vec{a},\ \vec{b},\ \vec{c}

v⃗×v⃗=0⃗\vec{v}\times\vec{v}=\vec{0}

a⃗×(a⃗+b⃗+c⃗)=0⃗\vec{a}\times(\vec{a}+\vec{b}+\vec{c}) = \vec{0}

a⃗×b⃗+a⃗×c⃗=0⃗\vec{a}\times\vec{b} + \vec{a}\times\vec{c} = \vec{0}

a⃗×b⃗=c⃗×a⃗\vec{a}\times\vec{b} = \vec{c}\times\vec{a}

b⃗×(a⃗+b⃗+c⃗)=0⃗\vec{b}\times(\vec{a}+\vec{b}+\vec{c}) = \vec{0}

−a⃗×b⃗+b⃗×c⃗=0⃗-\vec{a}\times\vec{b} + \vec{b}\times\vec{c} = \vec{0}

b⃗×c⃗=a⃗×b⃗\vec{b}\times\vec{c} = \vec{a}\times\vec{b}

c⃗×(a⃗+b⃗+c⃗)=0⃗\vec{c}\times(\vec{a}+\vec{b}+\vec{c}) = \vec{0}

c⃗×a⃗−b⃗×c⃗=0⃗\vec{c}\times\vec{a} - \vec{b}\times\vec{c} = \vec{0}

c⃗×a⃗=b⃗×c⃗\vec{c}\times\vec{a} = \vec{b}\times\vec{c}

a⃗×b⃗=c⃗×a⃗\vec{a}\times\vec{b} = \vec{c}\times\vec{a}

b⃗×c⃗=a⃗×b⃗\vec{b}\times\vec{c} = \vec{a}\times\vec{b}

c⃗×a⃗=b⃗×c⃗\vec{c}\times\vec{a} = \vec{b}\times\vec{c}

∴ a⃗×b⃗=b⃗×c⃗=c⃗×a⃗\therefore\ \vec{a}\times\vec{b} = \vec{b}\times\vec{c} = \vec{c}\times\vec{a}

Suppose a⃗×b⃗=0⃗⇒a⃗∥b⃗\text{Suppose } \vec{a}\times\vec{b}=\vec{0} \Rightarrow \vec{a} \parallel \vec{b}

a⃗=b⃗⇒c⃗=−2a⃗, ∣c⃗∣=2≠1 ✗\vec{a}=\vec{b} \Rightarrow \vec{c}=-2\vec{a},\ |\vec{c}|=2 \neq 1\ \text{✗}

a⃗=−b⃗⇒c⃗=0⃗ ✗\vec{a}=-\vec{b} \Rightarrow \vec{c}=\vec{0}\ \text{✗}

a⃗=(1,0,0)\vec{a}=(1,0,0)

b⃗=(−12, 32, 0),c⃗=(−12, −32, 0)\vec{b}=\left(-\frac12,\ \frac{\sqrt3}{2},\ 0\right),\quad \vec{c}=\left(-\frac12,\ -\frac{\sqrt3}{2},\ 0\right)

a⃗×b⃗=b⃗×c⃗=c⃗×a⃗=(0, 0, 32)\vec{a}\times\vec{b}=\vec{b}\times\vec{c}=\vec{c}\times\vec{a}=\left(0,\ 0,\ \frac{\sqrt3}{2}\right)

∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta

=sin⁡120∘=32≠0= \sin 120^\circ = \frac{\sqrt3}{2} \neq 0

(1) a⃗×b⃗=0⃗ ✗(3) a⃗×b⃗=a⃗×c⃗ ✗(1)\ \vec{a}\times\vec{b}=\vec{0}\ \text{✗} \qquad (3)\ \vec{a}\times\vec{b}=\vec{a}\times\vec{c}\ \text{✗}

(4) equal, so parallel — not perpendicular ✗(4)\ \text{equal, so parallel — not perpendicular}\ \text{✗}

(2) a⃗×b⃗=b⃗×c⃗=c⃗×a⃗≠0⃗ ✓(2)\ \vec{a}\times\vec{b}=\vec{b}\times\vec{c}=\vec{c}\times\vec{a}\neq\vec{0}\ \text{✓}

a⃗×b⃗=b⃗×c⃗=c⃗×a⃗≠0⃗\vec{a}\times\vec{b} = \vec{b}\times\vec{c} = \vec{c}\times\vec{a} \neq \vec{0}

Correct option: (2)\text{Correct option: } (2)

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